A particle of mass `m` is projected at an angle `alpha` to the horizontal with an initial velocity `u`. The work done by gravity during the time it reaches its highest point is

Options

A. (a) `u^2sin^2alpha`
B. (b) `(m u^2cos^2alpha)/(2)`
C. (c) `(m u^2sin^2alpha)/(2)`
D. (d) `-(m u^2sin^2alpha)/(2)`

Explanation

To find the work done by gravity on a particle of mass `m` projected at an angle `alpha` with an initial velocity `u` until it reaches its highest point, we can follow these steps:

### Step 1: Understand the motion of the particle
The particle is projected at an angle `alpha` with an initial velocity `u`. The initial velocity can be resolved into two components:
- Horizontal component: \( u_x = u \cos \alpha \)
- Vertical component: \( u_y = u \sin \alpha \)

### Step 2: Determine the highest point of the trajectory
At the highest point of its trajectory, the vertical component of the velocity becomes zero. The horizontal component remains unchanged since there are no forces acting in the horizontal direction.

### Step 3: Calculate the maximum height (h)
Using the kinematic equation for vertical motion, we can find the maximum height \( h \) reached by the particle:
\[ h = \frac{u_y^2}{2g} = \frac{(u \sin \alpha)^2}{2g} = \frac{u^2 \sin^2 \alpha}{2g} \]

### Step 4: Calculate the work done by gravity
The work done by gravity can be calculated using the formula:
\[ W = -mgh \]
Substituting the expression for \( h \):
\[ W = -mg \left( \frac{u^2 \sin^2 \alpha}{2g} \right) \]
The \( g \) cancels out:
\[ W = -\frac{mu^2 \sin^2 \alpha}{2} \]

### Step 5: Conclusion
Thus, the work done by gravity during the time it reaches its highest point is:
\[ W = -\frac{mu^2 \sin^2 \alpha}{2} \]

### Final Answer
The work done by gravity is \( -\frac{mu^2 \sin^2 \alpha}{2} \).

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